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Global Journal of Mathematical Sciences: Theory and Practical. ISSN 0974-3200 Volume 9, Number 2 (2017), pp. 103-116 © International Research Publication House http://www.irphouse.com

δωα-Closed Sets in Topological Spaces

S.Chandrasekar1, T. Rajesh Kannan2 and M.Suresh3

1,2Department of , Arignar Anna Government Arts college, Namakkal(DT),Tamil Nadu, India. 2Department of Mathematics, RMD Engineering College, Kavaraipettai, Gummidipoondi, Tamil Nadu, India.

Abstract In this paper we introduced new type of closed sets is called δωα-closed sets in the topological spaces. Moreover we discuss the relations between δωα- closed sets and already available various closed sets. Also we studied some properties and applications of δωα-closed, δωα- and δωα-. Key Words: δωα-closed sets, δωα-open sets , δωα- interior and δωα- closure. AMS subject classification: 54C55,54A05

1. INTRODUCTION In 1968 Velicko introduced δ-closed in Topological spaces [19]. Using δ-closed set several results introduced by many researcher. ωα-closed set[1] introduced by S. S. Benchalli,etal., in the year 2009. Since the advent of these types of notions, several author have been introduced interesting results. δg-closed set introduced by Dontchev[3](1999). In 1965 Njastad[10] introduced -open sets. The aim of the present paper is study the concept of δωα-closed sets and its various characterizations are investigated. Also, we further studied about δωα-interior δωα-closure in topological spaces.

104 S.Chandrasekar, T. Rajesh Kannan and M.Suresh

2. PRELIMINARIES Throughout this research paper (X, ) (or simply X) represent topological spaces ,For a A of X, cl(A), int(A) and Ac denote the closure of A, the interior of A and the of A respectively. Let us recall the following definition, which are useful in the sequel. Definition: 2.1 A subset A of (X,τ) is called if

1) Regular closed set [18] if cl(int(A))=A.

2) Semi closed set [6] if int(cl(A))) A

3) - closed set [10] if cl(int(cl(A))) A

4) δ-closed [19] if A = clδ(A), where clδ(A) = {x ∈X: int(cl(G)) A ≠ϕ, G∈ and x∈G}.

5) g-closed set [5] if cl(A) ⊆ U whenever A⊆ U , U is open in (X, τ) .

6) αg-closed set[7] if αcl(A)⊆ U whenever A⊆ U , U is open in (X, τ).

7) gα-closed set[8] if αcl(A)⊆ U whenever A⊆ U , U is α open in (X, τ).

8) δg-closed set [3] if δcl(A)⊆ U whenever A⊆ U , U is open in (X, τ).

9) gδ-closed set [4] if cl(A)⊆ U whenever A⊆ U , U is δ-open in (X, τ).

10) δg#-closed set[20] if δcl(A)⊆ U whenever A⊆ U , U is δ-open in (X, τ).

11) ω-closed set [16] if cl(A)⊆ U whenever A⊆ U , U is semi open in (X, τ).

12) ωα-closed set [1] if αcl(A) ⊆ U whenever A⊆ U , U is ω- open in (X, τ).

13) αω-closed set [12] if ω-cl(A) ⊆ U whenever A⊆ U , U is α-open in (X, τ).

14) δg*-closed set[17] if δcl(A) ⊆ U whenever A⊆U , U is g-open in (X, τ).

15) gωα-closed set [2] if αcl(A) ⊆ U whenever A⊆ U , U is ωα -open in (X, τ).

16) g*ωα -closed set[11] if cl(A) ⊆ U whenever A⊆ U , U is ωα -open in (X, τ).

17) δ (δg)*-closed set[9] if δcl(A) ⊆ U whenever A⊆ U ,U is δg-open in (X, τ). The complements of the above listed closed sets are their concern open sets.

3. δωα-CLOSED SETS In this section we introduce δωα-closed set in (X ,τ).

δωα-Closed Sets in Topological Spaces 105

Definition: 3.1

A subset A of a topological space (X, τ) is called δωα-closed set if clδ(A)⊆U whenever A⊆ U , U is ωα- in (X, τ) .the complement of δωα-closed set is called δωα-open set. Proposition 3.2 Every δ-closed set is δωα-closed. Proof: Assume that A is a δ- closed and Let U be any ωα-open set such that A ⊆ U.Here A is

δ-closed. A=clδ(A) for any subset A in (X, τ) .This implies that clδ(A)⊆ U. Here A is δωα-closed Set. Remark: 3.3 Converse of the above theorem need not be true from the following example. Let (X, τ) be a topological space with X= {a, b, c}and τ = {ϕ, X, {a}, {a, c}, {c}, {b, c}}. Here we take a subset A={c}, A is δωα closed but not δ-closed. Proposition 3.4 Every regular closed set is δωα-closed. Proof: Assume that A is a regular closed set and Let U be any ωα-open set such that A ⊆

U.Here A is regular closed set. This implies A is a δ- closed and A=clδ(A) for any subset A in (X, τ) .This implies that clδ(A)⊆ U . Here A is δωα-closed Set. Remark: 3.5 Converse of the above theorem need not be true from the following example. Let (X, τ) be a topological space with X= {a, b, c, d, e}and τ = {ϕ, X, {a}, {b}, {d}, {a, b}, {a, d}, {b, d}, {a, b, d} }Here we take a subset A={a,b,c,e}.A is δωα closed but not r-closed. Proposition 3.6 Every δωα-closed set is δg-closed. Proof: Consider, A is a δωα closed set and Let U be an open such that A⊆ U.We know that, every open set is ωα-open set. Therefore U is an ωα open set such that A⊆ U.Here A is δωα-closed set, it imply that clδ(A) ⊆ U. Hence A is δg-closed in (X, τ).

106 S.Chandrasekar, T. Rajesh Kannan and M.Suresh

Remark: 3.7 Converse of the above the theorem need not be from thefollowing example. Let (X, τ) be a topological space with X= {a,b,c} and τ = {ϕ, X, {a}, {a, c}, {c}, {b, c}} . Here we take a subset A={a,c}.A is δg closed not δωα-closed set in (X, τ). Proposition:3.8 Every δωα-closed set is αg-closed. Proof: Let A be any δωα-closed set in (X, τ) and V be an open set such that A ⊆ V. Therefore V is an ωα- open set such that A⊆ V .Since A is δωα-closed set, it implies that clδ(A) ⊆ V. But αcl(A) ⊆ clδ(A) for any A. Therefore . αcl(A)⊆V. V is an open set in (X, τ) .Hence A is αg-closed. Remark: 3.9 Converse of the above theorem need not be true from the following example, Let (X, τ) be a topological space with X= {a, b, c, d}and τ = {ϕ, X, {a}, {b}, {a, b}, {a, b, c}.Here wetake a subset A= {d}.A is αg-closed but not δωα- set in (X, τ) . Proposition 3.10 Every δωα-closed set is αω -closed. Proof: Let A be a δωα-closed and U be α-open set in (X, τ) such that A⊆ U. Since every α- open set is ωα- open set and A is δωα-closed, then ωcl(A) ⊆ cl(A) ⊆ clδ(A))⊆U Where U is ωα- open set. It implies that ωcl(A) ⊆U. Hence A is αω-closed in (X, τ). Remark: 3.11 Converse of the above the theorem need not be from the following example. Let (X, τ) be a topological space with X= {a,b,c,d} and τ = {ϕ, X, {a}, {b} {a, b}, {a, c}, {a, b, c}} . Here we take a subset A={a,b,c}.A is αω closed not δωα-closedset in (X, τ). Proposition 3.12 Every δωα-closed set is gωα-closed but converse need not be true.

δωα-Closed Sets in Topological Spaces 107

Proof: Let A be a δωα-closed and U be an ωα-open set in (X, τ) .such that A⊆ U. Since A is

δωα-closed, then clδ(A)) ⊆ U. But cl(A) ⊆clδ(A)) ⊆ U [4], it implies that cl(A) ⊆ U. Hence A is gωα-closed set. Remark 3.13 Converse of the above theorem need not be true from the following example, Let (X, τ) be a topological space with X= {a, b, c} and τ = {ϕ, X, {a}, {a, b}}. Here we take a subset A={c}.A is gωα-closed but not δωα-closed. Proposition 3.14 Every δωα-closed set is g*ωα-closed but converse need not be true. Proof: Let A be a δωα-closed and U be an ωα-open set in (X, τ) such that A⊆ U. SinceA is

δωα-closed, then clδ(A))⊆U.But αcl(A)⊆clδ(A))⊆U,it implies that αcl(A) ⊆U.Hence A is g*ωα-closed set. Remark 3.15 Converse of the above theorem need not be true from the following example, Let (X, τ) be a topological space with X ={a, b, c} and τ = {ϕ, X, {a}, {a, b}}. Here we take a subset A= {c}.A is g*ωα- closed but not δωα closed. Theorem 3.16 Every δωα-closed set is gα-closedset set but converse need not be true. Proof: Let A be a δωα-closed and U be an α-open set in (X, τ) .such that A⊆ U. Since A is

δωα-closed, then clδ(A)) ⊆ U. because every α-open set is ωα-open set in (X, τ) .But αcl(A) ⊆clδ(A)) ⊆ U , it implies that αcl(A)⊆U. Hence A is gα-closed set. Remark 3.17 Converse of the above theorem need not be true from the following example, Let (X, τ) be a topological space with X = {a, b, c, }andτ = {ϕ, X, {a}}. Here we take a subset A={c}.A is gα-closed but not δωα closed. Theorem 3.18 Every δωα-closed set is δg#-closed but converse need not be true.

108 S.Chandrasekar, T. Rajesh Kannan and M.Suresh

Proof: Let A be a δωα-closed and U be an δ-open set in (X, τ) such that A⊆ U. Since A is

δωα-closed, then clδ(A)) ⊆Uwhere U is ωα-open set .it implies that clδ(A)) ⊆ U. Hence A is δg#-closed set. Remark 3.19 Converse of the above theorem need not be true from the following example, Let (X, τ) be a topological space with X = {a, b, c, }and τ = {ϕ, X, {a}}. Here we take a subset A={c}.Ais δg#-closed but not δωα closed. Theorem 3.20 Every δωα-closed set is gδ-closed but converse is notneed be true. Proof: Let A be a δωα-closed and U be anδ-open set in (X, τ) such that A⊆ U. Since A is

δωα-closed, then clδ(A)) ⊆ U. But cl(A) ⊆clδ(A)) ⊆ U , it implies that cl(A) ⊆ U. Hence A is gδ-closed set. Remark 3.21 Converse of the above theorem need not be true from the following example, Let (X, τ) be a topological space with X= {a, b, c, }and τ = {ϕ, X, {a}}. Here we take a subset A={c}.A is gδ -closed but not δωα closed. Remark 3.22 δωα-closed set is independent with following closed sets are closed set ,α-closed set, ωα-closed set , δg* -closed set ,and δ(δg)* closed set above result we can prove by using followingexamples.

Example 3.23 Let (X, τ) be a topological space with X={a,b,c} with τ = { ϕ, { a }, {a ,b},X}. Here we take a subset A={ c }.A is closed, α-closed and ωα-closed set but not δωα- closed set. Example 3.24 Let (X, τ) be a topological space with X={a,b,c} with τ = { ϕ,{ a },{ b,c },X}. The subset A={ b } is δωα-closed set but not closed, α-closed and ωα-closed in (X, τ).

δωα-Closed Sets in Topological Spaces 109

Example 3.25 Let (X, τ) be a topological space with X={a,b,c,d} with τ ={ ϕ, {c}, {a,b},{c ,d},{a, b, c},X}. Here we take a subset A={b,c,d}.A is not δg*and ,δ(δg)* closed set but δωα-closed set. Example 3.26 Let (X, τ) be a topological space with X={a,b,c} with τ = { ϕ, { a },{ a,b },X}. Here we take a subset A ={ c}.A is δg* and δ(δg)* closed set but not δωα-closed set. Diagram-I Where A B represents A implies B but B does not implies A Where A B represents A independent B each other.

4. PROPERTIES AND CHARACTERIZATION OF 훅훚훂-CLOSED Theorem 4.1 The finite of δωα-closed sets is δωα-closed. Proof:

Let {Ai/ i =1,2 …n} be finite class of δωα-closed of a space (X,τ). Then for each ωα-open Set Ui in (X, τ) containing Ai, clδ(Ai)⊆Ui i∈ {1,2,...n}.Hence n n ⋃i=1 Ai⊆⋃i=1 Ui= V. Since arbitrary union of ωα-open sets in (X,τ) is also ωα -open Set in (X,τ), V is ωα -open in (X,τ). Also ,⋃i clδ(Ai ) =clδ(Ui Ai)⊆V. Therefore Ui, Ai is δωα-closed in (X,τ). 110 S.Chandrasekar, T. Rajesh Kannan and M.Suresh

Remark 4.2: Intersection of any two δωα-closed sets in (X,τ) need not be δωα-closed. Let (X, τ) be a topological space with X= {a, b, c}with τ = {ϕ, X, {a}, {c}, {a, c}, {b, c}}. Here we take two subsets {a,b} and{b,c} are δωα-closed sets but their intersection of two sets {b} is not δωα-closed. Theorem 4.3

Let A be δωα-closed Set of (X,τ). Then clδ(A)-A does not contain a non empty ωα- Closed set. Proof:

Suppose that A is δωα-closed. Let G be ωα-closed set contained in clδ(A)-A. Now GCis ωα-open set of (X,τ) Such that A⊆GC. Here A isδωα-closed Set of (X,τ). Thus C clδ(A) ⊆G . C C Therefore G⊆( clδ(A)) .Also G⊆clδ(A)-A. There G⊆(clδ(A)) ∩ clδ (A)=ϕ. Hence G=ϕ. Hence clδ(A)-A does not contain a non emptyωα-Closed set. Theorem 4.4 If A is ωα-open and δωα-closed subset of (X,τ) then A is δ-closed subset of (X,τ). Proof:

We know that A⊆clδ(A ).Here A is ωα-open and δωα-closed,its implies that clδ(A) ⊆A. Hence A is δ-closed. Theorem 4.5 The intersection of a δωα-closed set and a δ-closed set is δωα-closed. Proof: Let A be δωα-closed and Let G be δ-closed. Take U is an ωα-open set with A∩G⊆U. C C C This imply A⊆U∪G .Here U∪G is ωα-open set .Therefore clδ (A )⊆U∪G .Now C clδ(A∩G) ⊆clδ(A)∩clδ G= clδ(A)∩ G⊆ (U∪G ) ∩ G = U. Here A ∩F is δωα-closed. Theorem: 4.6

If A is a δωα-clsoed set is space (X,τ) and A⊆B⊆clδ(A) then B is also δωα-clsoed set. Proof: Let U be a ωα-open set of (X,τ) such that B⊆U. Then A⊆U. Since A is δωα-closed set, clδ(A) ⊆U. Since B⊆clδ(A), clδ(B) ⊆clδ(clδ(A))= clδ(A). Hence clδ(B) ⊆U. Therefore B is also a δωα-closed set.

δωα-Closed Sets in Topological Spaces 111

Theorem 4.7.

Let A be δωα-clsoed of (X,τ). Then A is δ-closed set if clδ(A)-A is ωα-closed. Proof:

Let A be a δ-closed Subset of X. Then clδ(A)=A and so clg(A)-A=ϕ which is ωα - closed.

Conversely ,Here A is δωα- closed by theorem 4.3 clδ(A)-A does not contain a non- empty ωα-closed set. This imply clδ(A)-A=ϕ. Therefore clδ(A)=A. Hence A is δ- closed.

5. δωα- CLOSURE AND δωα- INTERIOR IN TOPOLOGICAL SPACES In this section, we introduce the notion of δωα- Closure and δωα-Interior of topological Spaces. Definition 5.1. The δωα- closure of a Subset A of X is denoted by δωα-cl(A) and is defined as the intersection of all δωα-closed sets containing A.Therefore δωα-cl(A) = ∩ {F ⊆ X: A ⊆ F and F is δωα-closed set }.

Definition 5.2 The δωα-interior of subset A of X is denoted by δωα-(int A) and is defined as the union of all δωα-open sets contained in A . Therefore δωα-int(A)= ∪ {G ⊆X: G ⊆ A and G is δωα- open set}. Remark 5.3 If A⊆ X then,

(i) A⊆δωα-cl(A) ⊆clδ(A).

(ii) intδ(A) ⊆int (A) ⊆ A. Theorem: 5.4 Let A and B be any two subsets of a space X then the following properties are true. (i) A is δωα-closed set if and only of δωα-cl(A) = A. (ii) δωα-cl(A) is the smallest δωα- closed subset of X containing A. (iii) δωα- cl(ϕ) = ϕ and ϕδωα-cl(X) = X. (iv) δωα-cl (A) is a δωα- closed set in (X, τ) .. 112 S.Chandrasekar, T. Rajesh Kannan and M.Suresh

(v) If A ⊆B, then δωα-cl(A) ⊆δωα- cl(B) (vi) δωα-cl (A ∪B) = δωα-cl(A) ∪δωα-cl(B) (vii) δωα-cl(A∩B) ⊆δωα-cl(A) ∩ δωα-cl(B) (viii) δωα-cl(δωα-cl(A)) = δωα-cl(A)

Proof: (i) we know that A ⊆δωα-cl(A) for any subset A of X. Let A be a δωα- closed set in (X, τ) .Also A ⊆ A and A ∈{F ⊆X: A ⊆F and F is δωα- closed Set}, it implies that A= ∩{F⊆X: A ⊆F and F is δωα- closed Set} ⊆ A. Then δωα- cl(A) ⊆ A. . Hence A= δωα-cl(A).converse is true from the direct definition. (ii) By the definition of δωα-cl(A) ,the intersection of any collection sets is closed. There δωα-cl(A) is closed.Also if B is any δωα-closed set containing then δωα-cl(A)⊆B.Therefore δωα-cl(A)is the smallest δωα-closed set in(X, τ) containingA. (iii) and (iv) –it follows from the Definition . (v) We know that B ⊆δωα-cl(B)for every B. if A⊆ B then A ⊆δωα-cl(B) .So δωα-cl(B) is the δωα-closed set containing A. But δωα-cl(A) is smallest δωα- closed set containing A . Therefore δωα-cl(A) ⊆δωα-cl(B). (vi) We know that A⊆ A ∪ B and B ⊆ A ∪ B, applying (v) ,we get δωα- cl(A)⊆δωα-cl(A ∪B) and δωα-cl(B) ⊆δωα-cl(A ∪B). Then δωα-cl(A) ∪δωα- cl (B) ⊆δωα-cl (A∪B). onthe other hand, δωα-cl(A) is δωα- closed set containing A and δωα-cl(B) is δωα-closed set containing B. Therefore δωα- cl(A) ∪δωα-cl (B) is δωα-closed set containing A∪B. But δωα-cl(A∪B) is δωα-closed set containing A∪B.Thereforeδωα-cl(A) ∪δωα-cl (B)⊇δωα-cl(A ∪B).Therefore δωα-cl (A ∪ B) = δωα-cl (A)∪δωα-cl (B). (vii) We know that A ∩ B ⊆ A and A ∩ B ⊆ B. by using (v) we have, δωα-cl(A ∩B) ⊆δωα- cl(A) and δωα- cl(A∩B) ⊆δωα- cl(A) ⋂δωα- cl(B). (viii) we know that δωα-cl(A) is a δωα- closed set in (X, τ) .Let δωα- cl(A) = G, then G is δωα- closed Set in (X, τ).From (i) we have, δωα-cl(G) = G. It implies that δωα- cl(δωα- cl(A)) = δωα- cl(A).In the example3.5, subsets A= {a} and B= {c}, then δωα- cl(A)= {a,c}, δωα- cl(B)= {b,c} and δωα- cl(A ∩ B) = ϕ. Therefore {c} = δωα- cl(A) ∩ δωα-cl(B) ⊆δωα- cl(A∩B)= ϕ.

δωα-Closed Sets in Topological Spaces 113

Definition 5.5 Let A be any subset of a space X. Then, ωα- of A is denoted by ωα-kernel (A) and Is defined as intersection of all ωα open sets containing A. Then ωα-kernel(A) = ∩ {U ⊆X: A ⊆U and U is ωα open set}. Theorem 5.6

A subset of a space X is δωα-closed set if and only if clδ(A) ⊆ωαker(A). Proof:

Let U be a ωα-open set containing A then clδ(A)⊆ωαker(A) ⊆ U. Therefore A is δωα- closed set.

Conversely suppose A is δωα- closed in (X, τ) .Then clδ(A) ⊆ U where U is ωα-open in (X, τ).

Let x ∈clδ(A).if x does not belong to ωα-kernel(A) then there exists an ωα-kernel(A) set U containing A such that x does not belong to U. Then x does not belong to clδ(A), which is a contradiction to the hypotheses. Hence clδ(A) ⊆ωαker(A).

Definition 5.7 Let N be any subset of topological space X them N is said to be δωα- neighborhood (denoted by δωα-nbd) of point x ∈ X if there exist an δωα- open set U such that x ∈ U ⊆ N. Theorem 5.8 A subset A of topological space X is δωα- closed and x ∈δωα-cl(A) if and only if N ∩ A ≠ ϕ for Any δωα-nbd N of x in (X, τ) . Proof: Suppose x∉δωα-cl(A). Then there exists δωα-closed set F of X such that A ⊆ F and x ∉ F. Thus x ∈ (X-F) is δωα- open in X. But A ∩ (X-F) = ϕ which is a contradiction. Hence x∈δωαcl(A).Conversely, suppose that there exists an δωα-nbd N of a point x ∈ X such that N ∩ A = ϕ Then there exists an δωα-open set F of X such that x -open set F of X such that x∈ F ⊆ N. Therefore we have F ∩A = ϕ and x ∈ (X-F). Then δωα-cl(A) ∈ (X-F) and x ∉δωα-cl(A), which is a contradiction to hypothesis that x ∈δωα-cl(A). Therefore N ∩ A ≠ ϕ.

114 S.Chandrasekar, T. Rajesh Kannan and M.Suresh

Remark 5.9: The intersection of any two member of δωα-N(x) is again a member of δωα-N(x). Definition 5.10 Let a be a subset of topological space X. Then a point x ∈ X is said to be a δωα- of A if every δωα-oen set of x contains a point of a other than, x that is G ∩ (A- {x})≠ϕfor every δωα- open set G of X. In a topological space X, the set of all δωα- limit point of given a subset A of X is called δωα-derived set of A and is denoted by δωα-d(A). Theorem 5.11 Let A and B be any two subsets of a space X then the following properties are true: (i)δωαd(ϕ) = ϕ (ii) If A ⊆ B, then δωα-d(A) ⊆δωα-d(B) (iii) δωα-d(A ∪ B) = δωα-d(A) ∪δωα-cl(B) (iv) δωα-d(A ∩ B) ⊆δωα-d(A) ∩ δωα-d(B).

Proof: (i) Let x ∈ X and x ∈δωα-d(ϕ). Then for every δωα-open set G containing x, we should have G ∩ (A-{x})=ϕ, which is impossible. Therfore δωα-d(ϕ) = ϕ.

(ii) Let x ∈δωα-d(A) then x is a limit point of A, G ∩ (A-{x})≠ ɸ for every δωα-nbd G containing x. if A ⊆ B, then G ∩ (B-{x})≠ϕ. Therefore x ∈δωα-d(B). Hence δωα- d(A) ⊆δωα-d(B) (iii) We know that A⊆ A ∪ B and B ⊆ A ∪ B then from property (ii), δωα-d(A) ∪δωα-d(B) ⊆δωα-d(A ∪ B) .. On the other hand if x ∉δωα-d(A) ∪δωα-d(B), then x

∉δωα-d(A) and x ∉δωα-d(B). Therefore there exist δωα-nbds G1 and G2 of x such that

G1∩(A-{ x}) = ϕ and G2∩ (B-{ x })= ϕ. Since G1 ∩ G2 is δωα-nbd of x, then we get (G1 ∩ G2 ) ∩ [A∪B- { x }] = ϕ. Therefore x ∉δωα-d(A ∪ B). Therfore δωα-d(A ∪ B) ⊆δωα-d(A) ∪δωα-d(B) . we get, δωα-d(A ∪ B) = δωα-d(A) ∪δωα-cl(B). (iv)Since A ∩ B ⊆ A and A ∩ B ⊆ B, then from property (ii) we have δωα-d(A ∪ B) ⊆δωα-d(A) and δωα-d(A ∩ B) ⊆δωα-d(B).Consequently, δωα-d(A ∩B) ⊆δωα-d(A) ∩ δωα-d(B)

δωα-Closed Sets in Topological Spaces 115

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